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台灣加權股價指數價量分析

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3.2 台灣加權股價指數價量分析

探討報酬波動的行為一直是經濟學家和計量經濟學家們非常感興趣的研究。

本研究欲從模糊理論的觀點切入,並以區間迴歸分析架構,藉由價量分析預測價 的波動區間。本研究蒐集2015 年 6 月 1 日~ 2016 年 4 月 30 日台灣證券交易所 發佈的台灣加權股價指數價量資料,指數的價為自迴歸行為的變數,前一期的價 與一周的交易量為獨立變數,資料採取最大值與最小值的區間資料作為我們的模 糊樣本。

考慮區間迴歸模型如下:

𝐘價,𝐭= 𝐀𝟎+ 𝐀𝟏𝐗+ 𝐀𝟐𝐗價,𝐭−𝟏 (3.7) Y價,t為今日價格變化,以區間模糊數表示

X為交易量(前五天交易日)變化,以區間模糊數表示

X價,t−1為昨日價格變化,以區間模糊數表示

採用二維模糊迴歸方法,估算出中心迴歸模式與半徑迴歸模式如下:

(1)中心迴歸模式:

𝐘價,𝐭,𝐂= 𝟐𝟎𝟓 + 𝟎. 𝟗𝟕𝐂價,𝐭−𝟏+ 𝟎. 𝟏𝟒𝐂 (3.8)

(2)半徑迴歸模式:

𝐘價,𝐭,𝐒= 𝟑𝟒. 𝟏𝟏 + 𝟎. 𝟐𝟕𝐒價,𝐭−𝟏+ 𝟎. 𝟎𝟒𝐒 (3.9) 藉由中心迴歸模式與半徑迴歸模式,我們針對105 年 5 月 3 日~5 月 20 日的 台灣加權股價指數價格資料進行估計,其結果如下:

𝑓(< 0,1768.51; 823.02 >)

= 472.75 (3.10)

由結果顯示,估計值與實際值 14 天距離加總,最近距離總和為 0;最遠距 離總和為1768.51;重疊距離總和為 823.02;平均距離總和為 472.75。

接著我們將比較2.6 節(其它模糊迴歸方法),陳孝煒、吳柏林(2007),區

區間距離為:< 0,1582.79; 629.41 >

平均距離為:476.69

區間距離為:< 0,2693.88; 1179.95 >

平均距離為:756.97

(3) 乘法定義 2.5 之估計值與實際值距離加總

∑ D(Yi, Ŷ)i

14

i=1

= f(< 0,33634.08; 1260.3 >) = 16186.89

區間距離為:< 0,33634.08; 1260.3 >

平均距離為:16186.89

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由上述結果可知,定義 2.3 估計值與實際值區間平均距離加總最小,因此樣 本的最佳區間相乘定義為乘法定義 2.3。

圖3.7、圖 3.8 的結果中,明顯的觀察出區間乘法定義 2.4 與 2.5 所估計出來 的𝐘(𝐗𝐢)2.4、𝐘(𝐗𝐢)2.5與實際樣本值差異非常大。結果與3.1 節懸浮微粒PM10濃度 預測的結果相似,因此乘法定義2.4 與 2.5 與此樣本的配適度不佳。

由圖3.6 與圖 3.7,我們可看出乘法定義 2.3 與 2.4 兩著之間的差異,對於區 間半徑的預測。乘法定義2.4 預測過大,也造成較大的誤差,與乘法定義 2.3 相 反。

比較二維度技術與乘法定義2.3 的結果,兩者與 14 筆實際值資料皆有交集,

最小距離總和皆為零。雖然,最遠距離總和的結果中,二維度技術的誤差相較於 乘法定義2.3 來的差。但以覆蓋率(重疊距離)的結果,乘法定義 2.3 較不佳。

藉由區間數平均距離定義 2.2,同時考量三者因素(最近距離、重疊距離、

重疊距離),二維度技術在整體的表現中,較貼近樣本(平均距離最小)。

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對於隨時都在變動的觀測值,我們無法以單一數值精準預估。但藉由長期觀 察,可發現變動樣本週期性的規律。就像是週期波的形狀,具有一定的波峰跟波 谷,亦即數據變動範圍有上下界,超出上界或低於下界的機會很低。因此,如何 估計出上下界是很重要的一門技術,這是傳統迴歸較少考慮的問題。

傳統迴歸與時間序列都是預測下一刻的數值(橫向預測),然而在統計學中,

這樣的預測往往會因隨機性造成誤差。而模糊迴歸除了橫向預測外,同時進行縱 向預測(上下界預測)。藉由縱向的預測,我們可以做出相對應的決策。因此,

我們認為以模糊迴歸作為工具,配合我們提出之距離定義來選取適切之模型參數,

來表達與解釋許多生活中的樣本特性,可彌補傳統迴歸不足之處。

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6 附錄

Lemma 4 三角不等式:𝐃(𝐀

̃, 𝐂̃) ≤ 𝐃(𝐀̃, 𝐁̃) + 𝐃(𝐁̃, 𝐂̃) ∀𝐀̃, 𝐁̃, 𝐂̃ ∈ 𝐈𝐑

<pf> Suppose that A1+A2 2B1+B2 2C1+C2 2 .

This implies that MAB= B2− A1 , MBC = C2− B1 and MAC= C2− A1.

Case 1

C2 ≤ B2 ≤ A2

LAB= LBC= LAC= 0

CAB= −(B1− B2) CBC= CAC = −(C1− C2)

(B1−B2)+(B2 2−A1)+(C1−C2)+(C2 2−B1)= (C1−C2)+C2 2−A1

D(Ã, B̃) + D(B̃, C̃) = D(Ã, C̃)

Case 2

B2 ≤ C2 ≤ A2 且 A1 ≤ B1 ≤ C1 且 B2 ≥ C1 LAB= LBC= LAC= 0

CAB= −(B1− B2) CBC= −(C1− B2) CAC = −(C1− C2)

(B1−B2)+(B2 2−A1)+(C1−B2)+(C2 2−B1)(C1−C2)+C2 2−A1

D(Ã, B̃) + D(B̃, C̃) ≥ D(Ã, C̃)

Case 3

B2 ≤ C2 ≤ A2 且 A1 ≤ B1 ≤ C1 且 B2 ≤ C1 LAB= LAC = 0 LBC= C1− B2

CAB= −(B1− B2) CBC= 0 CAC = −(C1− C2)

(B1−B2)+(B2 2−A1)+(C1−B2)+(C2 2−B1)(C1−C2)+C2 2−A1

D(Ã, B̃) + D(B̃, C̃) ≥ D(Ã, C̃)

Case 4

B2 ≤ C2 ≤ A2 且 A1 ≤ C1 ≤ B1

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(B1−A2)+(B2 2−A1)+(C1−C2)+(C2 2−B1)(C1−A2)+C2 2−A1

D(Ã, B̃) + D(B̃, C̃) ≥ D(Ã, C̃)

Case 9

A2 ≤ C2 ≤ B2 且 A1 ≤ B1 ≤ C1 且 B1 ≥ A2 LBC= 0 LAB= (B1− A2) LAC= (C1− A2) CAB= 0 CBC= −(C1− C2) CAC = 0

(B1−A2)+(B2 2−A1)+(C1−C2)+(C2 2−B1)(C1−A2)+C2 2−A1

D(Ã, B̃) + D(B̃, C̃) ≥ D(Ã, C̃)

Case 10

A2 ≤ C2 ≤ B2 且 B1 ≤ A1 ≤ C1 且 A2 ≥ C1 LAB= LBC= LAC= 0

CAB= −(A1− A2) CBC= −(C1− C2) CAC = −(C1− A2)

(A1−A2)+(B2 2−A1)+(C1−C2)+(C2 2−B1)(C1−A2)+C2 2−A1

D(Ã, B̃) + D(B̃, C̃) ≥ D(Ã, C̃)

Case 11

A2 ≤ C2 ≤ B2 且 B1 ≤ A1 ≤ C1 且 A2 ≤ C1 LAB= LBC= 0 LAC= (C1− A2)

CAB= −(A1− A2) CBC= −(C1− C2) CAC = 0

(A1−A2)+(B2 2−A1)+(C1−C2)+(C2 2−B1)(C1−A2)+C2 2−A1

D(Ã, B̃) + D(B̃, C̃) ≥ D(Ã, C̃)

Case 12

A2 ≤ C2 ≤ B2 且 B1 ≤ C1 ≤ A1

LAB= LBC= LAC= 0

CAB= −(A1− A2) CBC= −(C1− C2) CAC = −(A1− A2)

(A1−A2)+(B2 2−A1)+(C1−C2)+(C2 2−B1)(A1−A2)+C2 2−A1

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(B1−B2)+(B2 2−A1)+(C1−B2)+(C2 2−B1)(C1−A2)+C2 2−A1

D(Ã, B̃) + D(B̃, C̃) ≥ D(Ã, C̃)

Case 22

A2 ≤ B2 ≤ C2 且 A1 ≤ B1 ≤ C1 且 A2 ≤ B1 且 B2 ≤ C1

LBC= (C1− B2) LAC= (C1− A2) LAB= (B1− B2) CAB= 0 CBC= 0 CAC = 0

(B1−B2)+(B2 2−A1)+(C1−B2)+(C2 2−B1)(C1−A2)+C2 2−A1

D(Ã, B̃) + D(B̃, C̃) ≥ D(Ã, C̃)

Case 23

A2 ≤ B2 ≤ C2 且 B1 ≤ A1 ≤ C1 且 A2 ≥ C1 LAB= LBC= LAC= 0

CAB= −(A1− A2) CBC= −(C1− B2) CAC = −(C1− A2)

(A1−A2)+(B2 2−A1)+(C1−B2)+(C2 2−B1)(C1−A2)+C2 2−A1

D(Ã, B̃) + D(B̃, C̃) ≥ D(Ã, C̃)

Case 24

A2 ≤ B2 ≤ C2 且 B1 ≤ A1 ≤ C1 且 A2 ≤ C1 ≤ B2 LAB= LBC= 0 LAC= (C1− A2)

CAB= −(A1− A2) CBC= −(C1− B2) CAC = 0

(A1−A2)+(B2 2−A1)+(C1−B2)+(C2 2−B1)(C1−A2)+C2 2−A1

D(Ã, B̃) + D(B̃, C̃) ≥ D(Ã, C̃)

Case 25

A2 ≤ B2 ≤ C2 且 B1 ≤ A1 ≤ C1 且 C1 ≥ B2 LAB= 0 LAC= (C1− A2) LBC= (C1− B2) CAB= −(A1− A2) CBC= 0 CAC = 0

(A1−A2)+(B2 2−A1)+(C1−B2)+(C2 2−B1)(C1−A2)+C2 2−A1

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LAB= LBC= LAC= 0

CAB= −(A1− A2) CBC= −(C1− B2) CAC = −(A1− A2)

(A1−A2)+(B2 2−A1)+(C1−B2)+(C2 2−B1)(A1−A2)+C2 2−A1

D(Ã, B̃) + D(B̃, C̃) ≥ D(Ã, C̃)

Case 31

A2 ≤ B2 ≤ C2 且 C1 ≤ B1 ≤ A1

LAB= LBC= LAC= 0

CAB= −(A1− A2) CBC= −(B1− B2) CAC = −(A1− A2)

(A1−A2)+(B2 2−A1)+(B1−B2)+(C2 2−B1)(A1−A2)+C2 2−A1

D(Ã, B̃) + D(B̃, C̃) ≥ D(Ã, C̃)

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